# Is Value At Risk additive?

I have computed the value at risk of 2 different commodities.

Assuming they have not correlated, can I just sum the two standalone VaR to get my overall portfolio's VaR ?

• I have VAR for commodity 1 and VAR for commodity 2. – RyanB Nov 23 '16 at 0:48
• What do you mean? Do you want to compute the VaR of holding 1 unit of both commodities? – SRKX Nov 23 '16 at 1:15
• Yes var of holding both – RyanB Nov 23 '16 at 1:16

The answer to your question is no. Value at Risk is not additive in the sense that $\text{VaR}(X+Y) \neq \text{VaR}(X) + \text{VaR}(Y)$. But I guess your question is more to aimed at finding a formula for your investments than to look at the property itself.

I think the only way to get a nice formula for this is to assume that both assets are:

• Normally distributed
• Have a mean equal to 0
• Are independent

Closed-Form value at risk for Normal variable

Mathematically the Value at Risk at a given level $\alpha$ is defined as:

$$\text{VaR}_\alpha(X) = \{ y ~ | ~ \mathbb{P}( X\leq y) = \alpha \}$$

If you can assume you variable $X$ is normally distributed such that $X \sim \mathcal{N}(\mu, \sigma^2)$, then you can re-express $X$ in terms of another variable $Z \sim \mathcal{N}(0,1)$: $X = \mu + \sigma Z$.

Using this, we know can rewrite the VaR definition as:

\begin{align} \text{VaR}_\alpha(X) &= \{ y ~ | ~ \mathbb{P}( \mu + Z\sigma\leq y) = \alpha\}\\ &= \left\{ y ~ | ~ \mathbb{P}\left( Z \leq \frac{ y- \mu}{\sigma} \right) = \alpha \right\}\\ &= \left\{ y ~ | ~ \Phi\left( \frac{ y- \mu}{\sigma} \right) = \alpha \right\}\\ \end{align}

where $\Phi(x)$ is the cumulative normal standard distribution function.

We can then find a closed-form formula to the value at risk of a normally distributed variable $X$:

$$\text{VaR}_\alpha(X) = \Phi^{-1}(\alpha) \cdot\sigma + \mu$$

Distribution of portfolio of two Normal variables

Now, let's assume you portfolio $Y$ holds two assets $X_1$ and $X_2$ (the two commodities in your example), which are uncorrelated ($\rho = 0$).

If you assume that both are normally distributed $X_1 \sim \mathcal{N}(\mu_1,\sigma_1)$ and $X_2 \sim \mathcal{N}(\mu_2,\sigma_2)$, then the we know that the portfolio can be expressed as:

\begin{align} Y &= wX_1 + (1-w)X_2\\ &= w(\mu_1 + \sigma_1 Z_1) + (1-w)(\mu_2 +\sigma_2 Z_2)\\ &= w\mu_1 + (1-w) \mu_2 + w\sigma_1 + w \sigma_1 Z_1 + (1-w) \sigma_2 Z_2 \end{align}

Hence, we know that:

$$\mathbb{E}(Y) = w\mu_1 + (1-w) \mu_2$$

and

$$\text{Variance}(Y) = \sigma_Y^2 = w^2 \sigma_1^2 + (1-w)^2 \sigma_2^2$$

because you assets are independent.

As we know, the sum of 2 normally distributed variables is also normally distributed, hence: $$Y \sim \mathcal{N}(w\mu_1 + (1-w) \mu_2, w^2 \sigma_1^2 + (1-w)^2 \sigma_2^2)$$

Value-at-risk of the portfolio

Using the formula for value-at-risk for normal variable we found above, we can write:

\begin{align} \text{VaR}_\alpha(Y) &= \Phi^-1(\alpha) \sigma_Y + \mu_y\\ \text{VaR}_\alpha(Y) &= \Phi^-1(\alpha) \sqrt{w^2 \sigma_1^2 + (1-w)^2 \sigma_2^2} + w\mu_1 + (1-w) \mu_2\\ \end{align}

If you assume that $\mu_1 = \mu_2 = 0$, then you get:

\begin{align} \text{VaR}_\alpha(Y) &= \Phi^-1(\alpha) \sqrt{w^2 \sigma_1^2 + (1-w)^2 \sigma_2^2}\\ \text{VaR}_\alpha(Y)^2 &= \Phi^-1(\alpha)^2 (w^2 \sigma_1^2 + (1-w)^2 \sigma_2^2)\\ \text{VaR}_\alpha(Y)^2 &= \Phi^-1(\alpha)^2 w^2 \sigma_1^2 + \Phi^-1(\alpha)^2 (1-w)^2 \sigma_2^2\\ \text{VaR}_\alpha(Y)^2 &= w^2 \text{VaR}_\alpha(X_1)^2 + (1-w)^2 \text{VaR}_\alpha(X_2)^2\\ \text{VaR}_\alpha(Y) &=\sqrt{ w^2 \text{VaR}_\alpha(X_1)^2 + (1-w)^2 \text{VaR}_\alpha(X_2)^2}\\ \end{align}

• Normal distribution is very rare situation in practice. Is it possible to extend your answer to general case? – Nick Nov 23 '16 at 7:23
• @Nick general case has no simple closed-form solution; it would depend on the joint distribution of the assets. – SRKX Nov 23 '16 at 7:36
• what is "w" in the above? – RyanB Nov 23 '16 at 12:58
• I'm guessing w is weight of % of each i own – RyanB Nov 23 '16 at 13:04
• Yes that's correct. And you VaR is expressed as a percentage of your initial wealth. – SRKX Nov 23 '16 at 14:10

You need to square them, add the squares , and take the square root. (Variances are additive, not standard deviations).

• That's it? I was reading about it violating sub additivity – RyanB Nov 23 '16 at 1:07
• VAR is the abbreviation of variation, VaR is the abbreviation value at risk. In the title you wrote value at risk, in the body you used VAR. What did you mean? – Nick Nov 23 '16 at 3:37
• Value at risk yes. Sorry about abbreviation. Value at risk in title. – RyanB Nov 23 '16 at 4:35
• @Nick I think VAR is more commonly used as Vector AutoRegression, though. – Jan Sila Nov 23 '16 at 8:31

Well, if you are using historical VaR, you can add results on each scenario and then calculate percentile of results... There is no other way.

No, because the value at risk is not, in general, a coherent risk measure as it does not respect the sub-additivity property, i.e.

$\rho(X + Y) \ne \rho(X) + \rho(Y)$, $\forall X, Y \in \mathcal{X}$ for the $VaR$.

However, Conditional Value at Risk is. Check out Is Conditional Value-at-Risk (CVaR) coherent?

• This is only showing it would not work in all cases, but you don't show why it couldn't work under certain assumption (here $\rho=0$). – SRKX Nov 23 '16 at 5:49